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DaniilM [7]
3 years ago
15

If you started the motor of a boat in the middle of a lake, who would detect the sound of the motor first: a friend sitting on t

he shore of the lake, or a friend snorkeling just below the surface of the water along the same shore? Explain your answer.
Physics
2 answers:
Andrei [34K]3 years ago
7 0

Answer:

A friend snorkeling just below the surface of the water along the same shore will detect the sound first.

ArbitrLikvidat [17]3 years ago
6 0

Answer:

A friend snorkeling just below the surface of the water along the same shore will detect the sound first.

Explanation:

  • The speed of sound in water medium is faster than that through the air.
  • Sound propagates through the medium by transferring through the molecules on it. Water has more closely packed molecules due to which the speed is faster.
  • In fact, the sound's speed in water is almost four times faster than that in the air.
  • So the guy in the water surface gets to hear sound faster than the one in sore.
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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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2 years ago
An unmarked police car traveling a constant 95 km/h is passed by a speeder. Precisely 1.00 s after the speeder passes, the polic
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Answer:

Speed of the speeder will be 28 m/sec

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In first case police car is traveling with a speed of 90 km/hr

We can change 90 km/hr in m/sec

So 90km/hr=\frac{90\times 1000m}{3600sec}=25m/sec

Car is traveling for 1 sec with a constant speed so distance traveled in 1 sec = 25×1 = 25 m

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So distance traveled by car in these 7 sec

S=ut+\frac{1}{2}at^2=25\times 7+\frac{1}{2}\times 2\times 7^2=224m

So total distance traveled by police car = 224 m

This distance is also same for speeder

Now let speeder is moving with constant velocity v

so 224=\left ( 1+7 \right )v

v = 28 m/sec

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